Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{y^2 - 13y + 42}{-7y - 42} \times \dfrac{y + 6}{-6y + 42} $
Answer: First factor the quadratic. $n = \dfrac{(y - 7)(y - 6)}{-7y - 42} \times \dfrac{y + 6}{-6y + 42} $ Then factor out any other terms. $n = \dfrac{(y - 7)(y - 6)}{-7(y + 6)} \times \dfrac{y + 6}{-6(y - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (y - 7)(y - 6) \times (y + 6) } { -7(y + 6) \times -6(y - 7) } $ $n = \dfrac{ (y - 7)(y - 6)(y + 6)}{ 42(y + 6)(y - 7)} $ Notice that $(y + 6)$ and $(y - 7)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(y - 7)}(y - 6)(y + 6)}{ 42(y + 6)\cancel{(y - 7)}} $ We are dividing by $y - 7$ , so $y - 7 \neq 0$ Therefore, $y \neq 7$ $n = \dfrac{ \cancel{(y - 7)}(y - 6)\cancel{(y + 6)}}{ 42\cancel{(y + 6)}\cancel{(y - 7)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $n = \dfrac{y - 6}{42} ; \space y \neq 7 ; \space y \neq -6 $